Time-Dependent Perturbation Theory

[PatsyTy]

Note this isn't actually a homework problem, I am working through my textbook making sure I understand the derivation of certain equations and have become stuck on one part of a derivation.

1. Homework Statement

I am working through my text (Quantum Mechanics 2nd Edition by B.H Bransden & C.J Joachain) section 9.1 on Time-Dependant Perturbation Theory. Part way through their derivation they give the equation

$i \hbar \sum_k \dot{c}_k (t)\psi_k^{(0)}e^{-\frac{iE_k^{(0)}t}{\hbar}}=\lambda \sum_kH'(t)c_k(t)\psi_k^{(0)}e^{-\frac{iE_k^{(0)}t}{\hbar}}$

where $c_k$ denotes the time-dependent coefficients, any factor with a $(0)$ superscript denotes a factor relating to the "unperturbed" state, $\lambda$ is just a constant (usually equal to one for physical systems) and $H'$ denotes the "perturbed" Hamiltonian.

From this equation the text states:

"Taking the scalar product with a function $\psi_b^{(0)}$ belonging to the set $y_k^{(0)}$ of unperturbed energy eigenfunctions, and using the fact that $\langle \psi_b^{(0)} | \psi_a^{(0)} \rangle = \delta_{bk}$, we find that

$\dot{c}_b (t) = i \hbar^{-1} \lambda \sum_k H'_{bk}(t)e^{i \omega_{bk} t} c_k(t)$

where $H'_{bk}=\langle \psi_{b}^{(0)} | H' (t) | \psi_k^{(0)} \rangle$ and $\omega_{bk}=\frac{E_b^{(0)}-E_k^{(0)}}{\hbar}$ is the Bohr angular frequency.

Homework Equations

Not sure if there are any not given really, it's mostly just rearranging everything

The Attempt at a Solution

So first I multiply both sides by $e^{\frac{iE_k^{(0)}t}{\hbar}}$ cancel it out on the L.H.S and bring all the big terms to the R.H.S giving

$i \hbar \sum_k \dot{c}_k (t)\psi_k^{(0)}=\lambda \sum_k c_k(t)\psi_k^{(0)}e^{\frac{iE_k^{(0)}t}{\hbar}}H'(t)e^{-\frac{iE_k^{(0)}t}{\hbar}}$

I then change all the $\psi_k^{(0)}$ eigenfunctions to their ket representation

$i \hbar \sum_k \dot{c}_k (t)| \psi_k^{(0)} \rangle=\lambda \sum_k c_k(t)e^{\frac{iE_k^{(0)}t}{\hbar}}H'(t)e^{-\frac{iE_k^{(0)}t}{\hbar}} | \psi_k^{(0)}\rangle$

I then multiply (i.e take the inner product) of the L.H.S and the R.H.S with $\langle\psi_b^{(0)}|$ giving

$i \hbar \sum_k \dot{c}_k (t) \langle\psi_b^{(0)}| \psi_k^{(0)} \rangle=\lambda \sum_k c_k(t)\langle\psi_b^{(0)}|e^{\frac{iE_k^{(0)}t}{\hbar}}H'(t)e^{-\frac{iE_k^{(0)}t}{\hbar}} | \psi_k^{(0)}\rangle$

The portion of this equation that is giving me issues is how the $k$ subscript in $\langle \psi_b^{(0)} | e^{\frac{iE_k^{(0)}t}{\hbar}}$ changes to $b$ giving me $\langle\psi_b^{(0)}|e^{\frac{iE_k^{(0)}t}{\hbar}}$. If I can get this I believe all the exponential terms are not operators so I can pull them out of the braket term to give me $H'_{bk}=\langle \psi_{b}^{(0)} | H' (t) | \psi_k^{(0)} \rangle$ and and then I can just combine the exponentials to give me the Bohr angular frequency term.

I'm sure I am doing something small (or something big) wrong somewhere but I just can't figure it out. Any help would be greatly appreciated!

Thanks!

 

[Fightfish]

So first I multiply both sides by $e^{\frac{iE_k^{(0)}t}{\hbar}}$ cancel it out on the L.H.S and bring all the big terms to the R.H.S giving

$i \hbar \sum_k \dot{c}_k (t)\psi_k^{(0)}=\lambda \sum_k c_k(t)\psi_k^{(0)}e^{\frac{iE_k^{(0)}t}{\hbar}}H'(t)e^{-\frac{iE_k^{(0)}t}{\hbar}}$

This is certainly one of the big wrong things - the exponential factor is part of the sum and is not a global factor, so you can't just multiply both sides by it.

 

[PatsyTy]

This is certainly one of the big wrong things - the exponential factor is part of the sum and is not a global factor, so you can't just multiply both sides by it.

Thanks, I see that now. Guess I'll be taking a look at it tomorrow and see if I can think of another way.

 

[vela]

Start with what the text says and take the scalar product. Also, remember that $H'(t)$ is an operator, so you can't move $\psi_k^{(0)}$ past it like you did on the RHS.

 

[PatsyTy]

Start with what the text says and take the scalar product. Also, remember that $H'(t)$ is an operator, so you can't move $\psi_k^{(0)}$ past it like you did on the RHS.

Thanks for the reply! I managed to figure it out. I was getting confused thinking $e^{-iE_k^{(0)}t/\hbar}$ was also an operator. When I get home I'll write out what I did to make sure it's correct.

Thanks again!

 

[PatsyTy]

So I believe the solution is to just change the eigenfunctions to their ket representation and multiply through by $\langle \psi_b^{(0)} |$

$i \hbar \sum_k \dot{c}_k (t)\langle \psi_b^{(0)} | \psi_k^{(0)} \rangle e^{-\frac{iE_k^{(0)}t}{\hbar}}=\lambda \sum_kc_k(t)\langle \psi_b^{(0)} | H'(t)| \psi_k^{(0)} \rangle e^{-\frac{iE_k^{(0)}t}{\hbar}}$

Which reduces to

$i \hbar \dot{c}_k (t) e^{-\frac{iE_b^{(0)}t}{\hbar}}=\lambda \sum_kc_k(t)H'_{ba} e^{-\frac{iE_k^{(0)}t}{\hbar}}$

then move the exponential over to the R.H.S and combine the terms for the Bohrn angular frequency to give what we want

$\dot{c}_k (t) =(i\hbar)^{-1}\lambda \sum_kH'_{bk} e^{-i\omega_{ba}t}c_k(t)$